The capacitive reactance is reduced by a factor of 2.
<h3>Calculation:</h3>
We know the capacitive reactance is given as,
where,
= capacitive reactance
f = frequency
C = capacitance
It is given that frequency is doubled, i.e.,
f' = 2f
To find,
=?
Therefore, the capacitive reactance is reduced by a factor of 2.
I understand the question you are looking for is this:
A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?
- The capacitive reactance is doubled.
- The capacitive reactance is traduced by a factor of 4.
- The capacitive reactance remains constant.
- The capacitive reactance is quadrupled.
- The capacitive reactance is reduced by a factor of 2.
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Answer:
He needs 1.53 seconds to stop the car.
Explanation:
Let the mass of the car is 1500 kg
Speed of the car, v = 20.5 m/s
He will not push the car with a force greater than, 
The impulse delivered to the object is given by the change in momentum as :
So, he needs 1.53 seconds to stop the car. Hence, this is the required solution.
Answer:
According to newton's second law of motionF=ma Data:-F=3200kgm/sec² or N ,a=2m/sec² ,m=? solution :-F=ma here we have to find m so m=F/a ,m=3200/2=1600kg
Answer:
10 gram gold is added
Explanation:
given data
pure gold = (24/24)
added = 15 grams
gold = 14 K ( 14/24)
gold = 18 K ( 18/24)
to find out
How much pure gold added
solution
we know here that when we add gold to get 20 K gold or 22K
so we added here 15 gram 14 K
we consider here m is pure gold added
so by composition here
we get
15 (14K) + m ( 24 K) = ( 15 + m ) (18) ...................1
solve it and find m
m = 10
so 10 gram gold is added
